2. Let us consider an integral π
π
π π₯ ππ₯ where f(x) be given certain equidistance value
of x, say π₯0 = π , π₯1 = π + β, β¦ β¦ . . , π₯ π = π + πβ = π. And the entries
corresponding to the arguments are π¦0, π¦1, π¦2, β¦ β¦ , π¦π respectively .
be a set of (n+1) values of the function y = f (x) corresponding to the
equidistant values π₯0, π₯1, π₯2, β¦ β¦ , π₯ π of the independent variable x.
Here π₯ π = π₯0 + πβ β β =
π₯ πβπ₯0
β
=
πβπ
β
where a is a lower bound of the interval
[a,b] and where b is the upper
bound of the interval [a,b] and n is the number of intervals.
6. Interpolation
β’ What is Interpolation?
β’ What is Extrapolation?
β’ Types
β’ Forward interpolation
β’ Backward interpolation
Y
X
7. Formula
β’ Newtonβs interpolation
β’ Forward difference interpolation formula
β’ Backward difference interpolation formula
β’ Lagrangeβs Interpolation formula
8. Newtonβs Formula for Interpolation
β’ Newtonβs Forward difference formula:
If the given data is
Then the newtonβs forward interpolation formula will be ,
π¦ π₯ = π¦0 + π’βπ¦0 +
π’(π’β1)
2!
β2 π¦0 +
π’(π’β1)(π’β2)
3!
β3 π¦0+β¦β¦β¦β¦..+
π’ π’β1 π’β2 β¦..2π’β¦..(πβ1)
π!
β π π¦0
Where u =
π₯βπ₯ π
β
h= difference of x which is always equal interval.
x π₯0 π₯1 π₯2 β¦β¦β¦β¦β¦ π₯ π
y π¦0 π¦1 π¦2 β¦β¦β¦β¦β¦. π¦ π
9. β’ Newtonβs Backward difference formula:
If the given data is
x π₯0 π₯1 π₯2 β¦β¦β¦β¦. π₯ π
y π¦0 π¦1 π¦2 β¦β¦β¦β¦β¦ π¦ π
Then the newtonβs Backward interpolation formula will be ,
π¦ π₯ = π¦π + π’π»π¦π +
π’(π’+1)
2!
π»2
π¦π +
π’(π’+1)(π’+2)
3!
π»3
π¦π+β¦β¦β¦β¦..+
π’ π’+1 π’+2 β¦..β¦..(π’+πβ1)
π!
π» π π¦0
Where u =
π₯βπ₯ π
β
h= difference of x which is always equal interval.
10. Example
d 50 55 60 65 70
A 1963 2376 2827 3318 3848
Find The area of circle of diameter 52.Where the area βAβ of circle of Diameter βdβ. Ans:2124
11. Lagrangeβs Interpolation Formula
β’ Given (n+1)Values of the Function f(x) forπ₯ = π₯0, π₯1,β¦β¦β¦β¦β¦β¦β¦. , π₯ π normally f(π₯0)
,f(π₯1) ,f(π₯2),β¦β¦β¦β¦ f(π₯ π) respectively the formula states:
π(π₯)=
(π₯βπ₯1)(π₯βπ₯2)β¦..(π₯βπ₯ π)
π₯0βπ₯1 (π₯0βπ₯2)β¦..(π₯0βπ₯ π )
π(π₯0)+
(π₯βπ₯0)(π₯βπ₯2)β¦..(π₯βπ₯ π)
π₯1βπ₯0 (π₯1βπ₯2)β¦..(π₯1βπ₯ π )
π(π₯1)+
(π₯βπ₯0)(π₯βπ₯1)(π₯βπ₯3)β¦β¦..(π₯βπ₯ π)
π₯2βπ₯0 (π₯2βπ₯1)(π₯2βπ₯3)β¦..(π₯2βπ₯ π)
π(π₯2)+
(π₯βπ₯0)(π₯βπ₯1)(π₯βπ₯2)β¦β¦..(π₯βπ₯ πβ1)
π₯ πβπ₯0 (π₯ πβπ₯1)(π₯ πβπ₯2)β¦..(π₯ πβπ₯ π+1)
π(π₯ π)
12. Example
X 321.0 322.8 324.2 325.0
πππ10 π₯ 2.50651 2.50893 2.51081 2.51188
Compute The value of πππ10323.5. π΄ππ : 2.50987
13. Comparisons Between Lagrange and Newtonβs Interpolation
Lagrange Newton
1.Lagrange method is numerically unstable 1.Newton's method is usually numerically stable and
computationally efficient.
2.the Lagrange formula less better for computation 2.Newton formula is much better for computation
than the Lagrange formula.
3.Lagrange's form is more efficient then the
Newton's formula when we have to interpolate
several data sets on the same data points
3.Less efficient for several data set.
14. Use of Interpolation in CSE
οComputer graphics.
οDrawing 2D Curves helps to find Bezier path in Adobe Illustrator , CorelDraw and Inkscape .
οComputer animation, interpolation is inbetweening.
οUse for 3D in laser light show .
ο Work as motion controller.
οUse to make Cartoon films.
οDefine the color of object, color location of frame.